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A fair coin is tossed 5 times. What is the probability that it lands

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A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post Updated on: 15 Mar 2019, 03:57
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A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

A. 1/16
B. 5/16
C. 2/5
D. 13/16
E. 27/32

Originally posted by netcaesar on 24 May 2006, 23:30.
Last edited by Bunuel on 15 Mar 2019, 03:57, edited 1 time in total.
Renamed the topic and edited the question.
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 24 May 2006, 23:44
5/16

Two heads can be in a toss of 5 times in 5C2 = 10 ways

Each way(number of times at least two heads will come) has a probability of 1/32.

Hence 10*1/32 = 5/16
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 24 May 2006, 23:45
1
It is B....

Total possible outcomes=2*2*2*2*2=32
Favourable outcomes=5C2=10

Probability=10/32=5/16....
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 24 May 2006, 23:52
1
SimaQ wrote:
It is B....

Total possible outcomes=2*2*2*2*2=32
Favourable outcomes=5C2=10

Probability=10/32=5/16....


going by your logic - the favourable outcomes should be 5C2 + 5C3 +5C4 + 5C5 ( question says at least 2..right?)
= 10 + 10 + 5 + 1 =26

hence prob = 26/32 = 13/16
ie choice D
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 25 May 2006, 00:02
iced_tea wrote:
SimaQ wrote:
It is B....

Total possible outcomes=2*2*2*2*2=32
Favourable outcomes=5C2=10

Probability=10/32=5/16....


going by your logic - the favourable outcomes should be 5C2 + 5C3 +5C4 + 5C5 ( question says at least 2..right?)
= 10 + 10 + 5 + 1 =26

hence prob = 26/32 = 13/16
ie choice D


Yes.......
Calculated ofr only two heads.....

Favorable outcomes are 5C2+5C3 + 5C4+5C5
=> 10+10+5+1
=> 26
26/32 = 13/16

Hence D
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 25 May 2006, 13:17
I don't understand whu you sum up "10+10+5+1".

Can someone explain it?
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 25 May 2006, 23:31
netcaesar wrote:
I don't understand whu you sum up "10+10+5+1".

Can someone explain it?


We sum up the number of ways you can toss heads up 2 times, 3 times, 4 times and 5 times....
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 06 Aug 2014, 20:43
1
Probability of all tails = 1/32
Probability of 4 tails = 5/32
------------------------------------------------------------------------------
Probability of all tails or 4 tails = 1/32 + 5/32 = 6/32 = (3/16)
------------------------------------------------------------------------------

Therefore probability of at least 2 heads = 1- Prob of all or 4 tails = 1- (3/16) = 13/16

Hence (D)
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 07 Aug 2014, 10:22
1
Fair coin is tossed 5 times . Hence total number of outcomes = 2^5 = 32.

Problem asks for probability of getting atleast heads twice. Hence if we calculate probability of getting Heads exactly once and probability of not getting Heads at all and subract it from the total probability of the event which is 1 (As total probability of certain event will be always 1) we can get the probability of getting atleast heads twice.

Probability of getting exactly one head and no heads= (Number of possible outcomes [ HTTTT , THTTT, TTHTT, TTTHT, TTTTH , TTTTT] = 6)/ (Total possible outcomes = 32)

=> 6/32 = 3/16

Hence probability of getting atleast heads twice = 1 - (3/16) = 13/16 => Choice [D]
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 17 Sep 2014, 07:14
netcaesar wrote:
A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

A. 1/16
B. 5/16
C. 2/5
D. 13/16
E. 27/32


Sol:

2H3T, 3H2T, 4H1T, 5H calculating this will take longer than calculating 5T, 4T1H

lets calculate 5T, 4T1H

5T= 1/2^5 = 1/32
4T1H = (1/32) (5!/4!) = 5/32

5T+4T1H= 6/32 = 3/16

1- (5T+4T1H) = 1-3/16 = 13/16

(D)
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 05 Jan 2017, 07:12
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Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.
If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 05 Jan 2017, 07:12
1
Here are the numbers to remember:

1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

The first row applies to 3 flip questions, the second to 4 flip questions and the third to 5 flip questions.
If we want at least 2 heads, we want 2 heads, 3 heads, 4 heads OR 5 heads. Pretty much whenever we see "OR" in probability, we add the individual probabilities.

Looking at the 5 flip row, we have 1 5 10 10 5 1. For 2H, 3H, 4H and 5H we add up the 3rd, 4th, 5th and 6th entries: 10+10+5+1=26.

Summing the whole row, we get 32.

So, the chance of getting at least 2 heads out of 5 flips is 26/32 = 13/16.
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 05 Jan 2017, 08:24
For all these kind of questions we can answer using this concept ( hope all of you might have studied during your school days):

Probability of 'r' success out of 'n' trials = = nCr (P)^r (Q)^n-r

P - probability of success
Q - probability of failure

According to this question :
We need to find probability of head >= 2 out of 5 trials.
P - probability of success = 1/2(probability of getting head out of a toss)
Q - probability of failure= 1/2(probability of getting tail out of a toss)

In mathematical form:
P(H>=2) = 1-P(H<2) = 1- (P( H=0) + P( H=1)).

P(H =0) = 5C0 (1/2)^0 (1/2)^5 = (1/2)^5---------- eqn 1

P(H =1) = 5C1(1/2)^1 (1/2)^4 = (5/2)^5---------- eqn 2

Finally,

1-P(H<2) = 1- (P( H=0) + P( H=1)) = 1 - ((1/2)^5)+(5/2)^5) = 1- (6 / 2^5) = 26/32= 13/16.


Final note, though this method looks lengthy, it will be applicable to all atleast,atmost and exactly equal to 'n' of questions in probability.

In GMAT, more than solving fast we have to keep our mind unstressed and fresh. So that we can be active for such a long exam.

Hope it helps [WINKING FACE]
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 18 Feb 2017, 09:15
Fair coin is tossed 5 times.total number of outcomes = 2^5 = 32

5C2 + 5C3 +5C4 + 5C5=10+10+5+1=26

26/32=13/16

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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 18 Feb 2017, 14:37
prob = 1 -(heads at max once)
= 1 - (heads exactly zero times or heads exactly one times)

at max once means two cases:
case 1 : exactly zero heads
=> TTTTT
=> 1/32

case 2 : heads exactly once
=> HTTTT
=> (5!/4!)/32
=> 5/32

Now after putting the values
=> 1 -(heads at max once)
=> 1 - ((1/32) + (5/32))
=> 13/16
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 18 Feb 2017, 22:10
We will have the following cases: (2H, 3T); (3H, 2T); (4H, 1T); (5H)

Also, HHHTT is different from HTTHT. Therefore, for (2H, 3T) => (1/2)^5 * 5!/(3! * 2!) = 10/32

(3H, 2T) => 10/32 ; (4H, 1T) => 5/32 ; (5H) => 1/32

Total = 13/16. Option D.
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 28 Mar 2018, 10:30
netcaesar wrote:
A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

A. 1/16
B. 5/16
C. 2/5
D. 13/16
E. 27/32


We can use the equation:

P(the coin lands heads up at least twice) = 1 - [P(it lands heads up exactly once) + P(it lands heads up 0 times)]

Let’s first compute the probability that the coin lands heads up exactly once:

P(HTTTT) = (½)^1 x (½)^4 = (½)^5 = 1/32

However, since HTTTT can be permuted 5!/(1! x 4!) = 5!/4! = 5 times, P(the coin lands heads up exactly once) = 5 x 1/32 = 5/32.

Now, let’s compute the probability that the coin lands heads up 0 times:

P(the lands heads up 0 times) = P(TTTTT) = (½)^5 = 1/32

Therefore,

P(the coin lands heads up at least twice) = 1 - [5/32+ 1/32] = 32/32 - 6/32 = 26/32 = 13/16

Answer: D
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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New post 10 Apr 2018, 02:18
netcaesar wrote:
A fair coin is tossed 5 times. What is the probability that it lands heads up at least twice?

A. 1/16
B. 5/16
C. 2/5
D. 13/16
E. 27/32


It is more easy way to with complement (at least 2 heads):

total - one head + no head
total = 2^5 = 32

one head = 5C1 + 5C0 = 5 + 1 = 6

Use complement here = 32 - 6 = 26/32 = 13/16.
I hope you understand this is most easiest way to do ;) ;)
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Re: A fair coin is tossed 5 times. What is the probability that it lands  [#permalink]

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